Descartes’ method for normals

11th March 2014

Descartes’ judgement on his method

The method: the elements

AM = x,CM = y, PA = v, PC = s.

Letf(x, y) = 0 (1)

be the equation of the curve.

The equation of the circle:

(v − x)2 + y2 = s2 (2)

.

The equation obtained by the elimination of x or y from (1) and(2) must have a double root.

Descartes’ method of indeterminate coefficients at workThe parabola: an example not given by Descartes

The example is chosen by Montucla.

y2 = 2px, (v − x)2 + y2 = s2.

x2 + 2(p− v)x + v2 + s2 = 0. We have to compare this equationwith

(x− e)2 = 0. Then p− v = −e, v = p + e.

But e = x⇒ v = p + x.

An example of Descartes

AM = x,CM = y,AP = v

y2 = rx− rqx

2 (Apollonius, I.13), (v − x)2 + y2 = s2.

The elimination of y2 and the comparison with (x− e)2 = 0 gives

v =1

2r + x− r

qx.

The tangent to the Conchoid

Descartes observes that is not very easy to determine the tangentto the conchoid by using his method. Nevertheless . . .

Van Schooten’s solution

The figure used by van Schooten:

GA = b, AE = LC = c, AM = BC = y, AB = x. v is taken onthe line EA.

AP = v, CP = s, and

x2 = s2 − (y + v)2.

The equation GM : MC = GA : AL

(b + y) : x = b : (x−√

c2 − y2). It is quite easy to obtain

x2y2 = (b + y)2(c2 − y2).

The elimination of x gives

y3 +c2 + v2 − b2 − s2

2v − 2by2 +

bc2

v − by +

b2c2

v − b= 0.

We have to compare the equation

y3 +c2 + v2 − b2 − s2

2v − 2by2 +

bc2

v − by +

b2c2

v − b= 0

with

(y − e)2(y − f) = 0.

It is very easy to get

v = b +bc2

y2+

b2c2

y3.

We have to prove Descartes’ construction.

v = b + bc2

y2+ b2c2

y3.

BCL, AGL are similar GL = bcy .

If DF = GL and CD = CB, from the similitude of CDF andCLH we have LH = bc2

y2. From the similarity of CDF and HIP

we have IP = b2c2

y3.

By AG, GI = LH, IP we can obtainAP = AG + GI + IP = b + bc2

y2+ b2c3

y3= v.